The product of numbers
3^(1/3).3^(1/9).3^(1/27)...... To infinity is
A.3
B.3^(1/3)
C.3^(1/2)
D.3√3
I am titally lost in this question and have absolutely no idea. Is there some rule or particular formula of solving such questions?
the exponents form a converging geometric series
multiplying (product) numbers with the same base (3 in this case) means adding the exponents
3^(1/3 + 1/9 + 1/27 + ...)
there is a formula for the sum of an infinite geometric series
... look it up
and the told riddle of moving to the door, 1/2 way, the 1/4 way, then 1/8 way...does she ever reach the door?
To find the product of the given sequence, let's simplify each term:
3^(1/3) = √(3^1) = √3
3^(1/9) = √(3^(1/3)) = (√3)^(1/3) = (3^(1/2))^(1/3) = 3^(1/6)
3^(1/27) = √(3^(1/9)) = (√3)^(1/9) = (3^(1/2))^(1/9) = 3^(1/18)
If you continue this pattern, you'll notice that each term can be written as 3 raised to a fractional exponent in the form 1/2^n. Now, let's express the general term in the sequence:
3^(1/2^n)
To find the product of an infinite sequence, we need to take the limit as n approaches infinity. As n approaches infinity, the exponent 1/2^n becomes smaller and smaller, eventually approximating zero.
Thus, the product of the given sequence is:
Product = lim (n → ∞) 3^(1/2^n)
Since 1/2^n approaches zero as n approaches infinity, the exponent in the limit becomes 3^0, which is equal to 1.
Therefore, the product of the given sequence is 3^1 = 3.
Hence, the answer is A. 3.